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NCERT Solutions Ganita Prakash (Part 1) Chapter 1 Figure it Out — Square Numbers

Book page 10 Updated on2026-09-05

Q1.
Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Answer

(i) 2032, (ii) 2048 and (iii) 1027 are not perfect squares. 1089 is.

NumberReasonVerdict
(i)2032ends in 2not a square
(ii)2048ends in 8not a square
(iii)1027ends in 7not a square
(iv)10891089 = 3 × 3 × 11 × 11 = 332a perfect square
Why the digit test suffices for the first three: No square ends in 2, 3, 7 or 8, so those three are ruled out on sight. The fourth ends in 9, which is allowed, so it needs a real check — and factorising gives 1089 = 332, so √1089 = 33.
Tip: 2048 = 211. The exponent 11 is odd, which is a second reason it cannot be a square.
Q2.
Which one among 64², 108², 292², 36² has last digit 4?
Answer

Only the units digit of the base matters.

SquareUnits digit of baseIts squareLast digit
6424166
10828644 ✓
2922244 ✓
3626366

Two of them qualify: 1082 and 2922.

Check it yourself: 1082 = 11664 and 2922 = 85264 — both end in 4; 642 = 4096 and 362 = 1296 end in 6. The question says "which one", but the numbers give two answers, so both must be named.
Why it happens: A square ends in 4 exactly when its root ends in 2 or 8, since 2 × 2 = 4 and 8 × 8 = 64 are the only single-digit products ending in 4.
Q3.
Given 125² = 15625, what is the value of 126²? (i) 15625 + 126 (ii) 15625 + 262 (iii) 15625 + 253 (iv) 15625 + 251 (v) 15625 + 512
Answer

(iv) 15625 + 251

1262 = (125 + 1)2 = 1252 + 2 × 125 + 1
= 15625 + 250 + 1 = 15625 + 251
= 15876
Why it happens: Consecutive squares differ by the next odd number, and here that is the 126th odd number, 2 × 126 – 1 = 251. Equivalently, 1262 – 1252 = (126 + 125)(126 – 125) = 251 × 1.
Check it yourself: 1262 = 126 × 126 = 15876, and 15625 + 251 = 15876. ✓
Q4.
Find the length of the side of a square whose area is 441 m².
Answer

The side is √441.

441 = 3 × 3 × 7 × 7
= (3 × 7) × (3 × 7) = 212
Side = 21 m
Why prime factorisation is the neat route: Both primes appear twice, so the factors split into two identical groups of 3 × 7. The product of one group, 21, is the square root. Estimating would also get you there — 202 = 400 and the number ends in 1, so the root ends in 1 or 9, and 21 is the only candidate in range.
Q5.
Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Answer

First find the LCM, then top it up to a square.

4 = 22, 9 = 32, 10 = 2 × 5
LCM = 22 × 32 × 5 = 180
In 180 the prime 5 appears once — an odd number of times
Multiply by 5: 180 × 5 = 900 = 22 × 32 × 52 = 302
Smallest such square = 900
Why 900 is smallest: Any number divisible by 4, 9 and 10 must be a multiple of their LCM, 180. A multiple of 180 that is a square must contain every prime of 180 an even number of times, so it needs at least one more 5. The least such multiple is 180 × 5 = 900, and 900 = 302 is indeed divisible by 4, 9 and 10.
Q6.
Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Answer

Factorise 9408 and look for the prime with an odd exponent.

9408 = 2 × 4704 = 2 × 2 × 2352 = 2 × 2 × 2 × 1176
= 2 × 2 × 2 × 2 × 588 = 2 × 2 × 2 × 2 × 2 × 294 = 2 × 2 × 2 × 2 × 2 × 2 × 147
147 = 3 × 49 = 3 × 7 × 7
So 9408 = 26 × 3 × 72

The 2s and the 7s pair up; only the single 3 is stranded. So multiply by 3.

9408 × 3 = 28224 = 26 × 32 × 72
= (23 × 3 × 7)2 = 1682
√28224 = 168
Why 3 is the smallest choice: To make every exponent even, each prime with an odd exponent must be supplied once more. Here that is only the prime 3, so the least multiplier is 3 itself; any smaller factor would leave the 3 unpaired.
Q7.
How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100
Answer

Between n2 and (n+1)2 there are 2n numbers.

(i) n = 16 → 2 × 16 = 32 numbers (from 257 to 288, since 162 = 256, 172 = 289)
(ii) n = 99 → 2 × 99 = 198 numbers (from 9802 to 9999, since 992 = 9801, 1002 = 10000)
Why it happens: (n + 1)2 – n2 = 2n + 1 counts the whole step from one square to the next. Excluding the square at the top end leaves 2n numbers strictly in between.
Check it yourself: 289 – 256 – 1 = 32 and 10000 – 9801 – 1 = 198. ✓
Q8.
In the following pattern, fill in the missing numbers: 1² + 2² + 2² = 3², 2² + 3² + 6² = 7², 3² + 4² + 12² = 13², 4² + 5² + 20² = (___)², 9² + 10² + (___)² = (___)²
Answer

Fourth row first:

42 + 52 + 202 = 16 + 25 + 400 = 441 = 212

Now read the rule from the completed rows. In each one the third base is the product of the first two, and the answer is one more than that product.

nn + 1n(n + 1)n(n + 1) + 1
1223
2367
341213
452021
9109091
92 + 102 + 902 = 81 + 100 + 8100 = 8281 = 912
Why the rule holds: Put k = n(n + 1). Then
n2 + (n+1)2 + k2 = n2 + n2 + 2n + 1 + k2 = 2n2 + 2n + 1 + k2 = 2k + 1 + k2 = (k + 1)2,
using 2n2 + 2n = 2n(n + 1) = 2k. So the identity works for every n, not just the rows shown.
Q9.
How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
Answer

Count in two stages — blocks, then tiny squares inside a block.

The picture is a 9 × 9 arrangement of blocks → 81 blocks
Each block (upright or tilted) is a 4 × 4 grid → 16 tiny squares
Total = 81 × 16 = 1296 tiny squares

Prime factorisation:

81 = 34, 16 = 24
1296 = 24 × 34
= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
Why the answer is a square — and more: Every prime here has an even exponent, so 1296 is a perfect square: 1296 = (22 × 32)2 = 362. Both exponents are also multiples of 4, so 1296 = 64 as well — it is a square of a square, which the picture shows directly: a square of blocks, each of which is a square of tiny squares.
Tip: Tilting a block does not change how many tiny squares it holds, so the diamonds count exactly like the upright ones.
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